some further chapter materials · some further chapter materials part one prologue to part one 1...

SOME FURTHERCHAPTER MATERIALS

PART ONE

Prologue to Part One1 Motion Matters2 Moving the Earth3 Understanding Motion4 Newton’s Unified Theory5 Conserving Matter and Motion6 The Dynamics of Heat7 Heat—A Matter of Motion8 Wave Motion9 Einstein and Relativity Theory

CHAPTER 1. MOTION MATTERS

Instantaneous Speed

In Section 1.3 we discuss the average speed of an object, which is definedto be the ratio of the distance traveled, �d, divided by the time interval, �t:

vav � .

You may have noticed that we cannot measure the speed of an object inan instant of time. The average speed is the only kind of speed that we canactually measure, since we can only measure distance intervals and time in-tervals. We can use more sophisticated instruments to obtain the distancetraveled in smaller and smaller time intervals. If the time interval �t has ap-proached zero, we are dealing with an instant in time, and the average speed

�d��t

25

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becomes the actual speed at that instant. This is called the instantaneous speed.However, in any real experiment we can never actually achieve an instantin time, an infinitesimally small time interval, since every measurement, nomatter how fast we can make it, still takes some amount of time.

Nevertheless, we can use a graph of the motion to calculate a reasonablevalue for the instantaneous speed at an instant of time. We point out inSection 1.4 that the slope of the line on a distance–time graph is

slope of line � ,

which is just the average speed during the time interval �t. As the time in-terval becomes smaller and smaller, the line on the graph during the timeinterval becomes straighter and straighter. In such a situation, for very tinytime intervals, the average speed becomes, by definition, equal to the in-stantaneous speed at the center of the time interval. To put it differently,as the value of �t approaches the limit of zero (which we cannot actuallymeasure), the value for the average speed vav approaches the instantaneousspeed, which is given the symbol v. In this case, the slope of the line be-comes a tangent to the curve at that instant. This means: the instantaneousspeed of an object at an instant of time t is defined as the tangent at time t to theline representing the object’s motion on a distance–time graph.

This can also be expressed in mathematical symbols as follows:

lim�t�0

� v.

In words, this says that in the limit as the time interval approaches zero,the ratio of the distance traveled divided by the time interval approachesthe instantaneous speed at the time t at the center of the original time interval. (Readers who have had some calculus may recognize this as a differential.)

Derivation of Galileo’s Expression d � 1⁄2at2

Galileo’s famous expression gives the distance (d ) traveled by an object start-ing from rest and moving with uniform acceleration (a) during the time in-terval (t). Note that this expression does not contain the speed, only thedistance and time, starting from zero, and the acceleration.

Galileo originally used a geometrical argument to derive this expression.Algebra was used more than 100 years later to derive the same expression.

�d��t

�d��t

26 SOME FURTHER CHAPTER MATERIALS

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Since it is more straightforward, we will use the algebraic derivation, alongwith some of Galileo’s original assumptions.

We start with the definition of the average speed of a uniformly accel-erating object during the time interval �t. (This expression holds no mat-ter how the object is moving.)

vav � .

We can rewrite this equation as

�d � vav � �t.

What would be the average velocity for a uniformly accelerating object?Galileo reasoned (as others had before him) that for any quantity thatchanges uniformly, the average value is just halfway between the beginningvalue and the final value. For uniformly accelerated motion starting fromrest, the initial speed is zero, vinitial � 0. So, the average speed is halfwaybetween 0 and vfinal:

vav � 1⁄2vfinal.

Substituting, we have

�d � 1⁄2vfinal � �t.

Now we have to obtain a value for vfinal. We can do this by starting withGalileo’s definition of average acceleration

aav � .

In our case, aav has a constant value, a, since the acceleration is uniform(constant). The value of �v is vfinal � vinitial, which is just vfinal, since vinitial � 0. Substituting in the equation for aav, we have

a � .

Rearranging, we get

vfinal � a � �t.

vfinal�

�t

�v��t

�d��t

PART ONE 27


So now we can replace vfinal in the expression for �d, and we obtain

�d � 1⁄2vfinal � �t,

�d � 1⁄2a(�t)2.

This is equivalent to Galileo’s expression. If we measure the distance andthe time interval from the position and the instant when the motion starts,then dinitial and tinitial are zero. We can then write this equation as

dfinal � 1⁄2atfinal2.

Or, if we let dfinal � d and tfinal � t, we have an even simpler expression

d � 1⁄2at2.

If we start with a nonzero initial speed, then we have

d � vinitial t � 1⁄2at2.

CHAPTER 3. UNDERSTANDING MOTION

Derivation of the Parabolic Trajectory of a Projectile

The motion of a projectile is composed of two independent motions: uni-form velocity in the horizontal direction and uniform acceleration in thevertical direction. During the time interval t, the distance traveled by theprojectile in the horizontal direction, x, with uniform speed vx is

x � vxt.

The distance the projectile moves in the vertical direction, y, during thesame time interval t is

y � 1⁄2gt2.

Solving the equation x � vx t for t gives

t � .x

�vx



Because the time interval t is the same in both equations, we can substitutex/vx for t in the equation for y. This gives

y � 1⁄2g� �2,

or

y � ��2v

g

x2�� x2.

This last equation contains two variables, x and y. It also contains threeconstant quantities: g, 2, and the horizontal speed vx. The vertical distancey that the projectile falls is thus a constant times the square of the hori-zontal displacement x:

y � (constant)x2.

The mathematical curve represented by this relationship between x andy is called a parabola. Galileo deduced the parabolic shape of trajectoriesby an argument similar to the one used here. This discovery greatly sim-plified the study of projectile motion, because the geometry of the parabolahad been established centuries earlier by Greek mathematicians.

x�vx

PART ONE 29

e

h

f i

d c

Drawing of a parabolic trajectoryfrom Galileo's Two New Sciences.

b aog

l

n

Derivation of the Equation for CentripetalAcceleration, ac � v2/R

Assume that a stone on the end of a string is moving uniformly in a circleof radius R. You can find the relationship between ac, v, and R by treatinga small part of the circular path as a combination of tangential motion andacceleration toward the center. To follow the circular path, the stone must


accelerate toward the center through a distance h in the same time that itwould move through a tangential distance d. The stone, with speed v, wouldtravel a tangential distance d given by d � v�t. In the same time �t, thestone, with acceleration ac, would travel toward the center through a dis-tance h given by h � 1⁄2ac �t2. (You can use this last equation because at t � 0, the stone’s velocity toward the center is zero.)

You can apply the Pythagorean theorem to the triangle in the figure thatfollows:

R2 � d2 � (R�h)2

� R2 � 2Rh � h2.

When you subtract R2 from each side of the equation, you are left with

d2 � 2Rh � h2.

You can simplify this expression by making an approximation. Since h isvery small compared to R, h2 will be very small compared to Rh. And since�t must be vanishingly small to get the instantaneous acceleration, h2 willbecome vanishingly small compared to Rh. So you can neglect h2 and write

d2 � 2Rh.

Also, d � v�t and h � 1⁄2ac�t2; so you can substitute for d2 and for h ac-cordingly. Thus,

(v �t)2 � 2R � 1⁄2ac (�t)2,

v2(�t)2 � Rac(�t)2,

v2 � Rac,

or

ac � �vR

2

�.

h

R

d

R



The approximation becomes better and better as �t becomes smaller andsmaller. In other words, v2/R gives the magnitude of the instantaneous cen-tripetal acceleration for a body moving on a circular arc of radius R. Foruniform circular motion, v2/R gives the magnitude of the centripetal ac-celeration at every point of the path. (Of course, it does not have to be astone on a string. It can be a small particle on the rim of a rotating wheel,or a house on the rotating Earth, or a coin sitting on a rotating phono-graph disk, or a car in a curve on the road, an electron in its path througha magnetic field, or the Moon going around the Earth in a nearly circularpath.)

The relationship among ac, v, and R was discovered by the Dutch sci-entists Christiaan Huygens and was published by him in 1673. Newton,however, must have known it in 1666, but he did not publish his proof un-til 1687, in the Principia.

We can substitute the relation v � 2�Rf or v � 2�R/T (see Section 4.11)into the equation for ac:

ac � �vR

2

�

�

� 4�2Rf 2

or

ac � .

These two resulting expressions for ac are entirely equivalent.

CHAPTER 4. NEWTON’S UNIFIED THEORY

“Weighing the Earth”

Now that we know how g arises in terms of Newton’s law of universal grav-itation, we can use the last equation above to find the mass of the Earth.This is possible because all of the terms in this equation are known, except

4�2R�

T 2

(2�Rf )2

�R

PART ONE 31


for MEarth. To find MEarth, first solve for it in the equation, using simple algebra:

MEarth � �gR

GEarth�.

(Be sure that you understand each step in obtaining the answer below; lookat the review of scientific notation in the Student Guide, if necessary.)

Now substitute the known values on the right side of the equation

MEarth � .

To obtain a result from this expression, we perform all of the indicatedarithmetic on the numbers and, separately, on the units. We’ll first collecteach of these together, which results in the following:

MEarth ��(9

6.8.6)(76.

�

4 �

101�

011

6)2

� .

For simplicity, let’s first work on the numbers (but never forgetting the units, which we’ll carry along). We start by squaring the term in thenumerator

(6.4 � 106)2 � 40.96 � 1012 �(Nm/

ms2

2

)/(mkg

)2

2

�.

So now we have

�(Nm/

ms2

2

)/(mkg

)2

2

�.

Multiply and divide the numbers, then subtract the exponent of the de-nominator from that of the numerator

MEarth � �(9.8

6)(.4607.96)

� � 1012 � 1011 �(Nm/

ms2

2

)/(mkg

)2

2

�

� 60.18 � 1023 � 6.02 � 1024 �(Nm/

ms2

2

)/(mkg

)2

2

�.

(9.8)(40.96 � 1012)��

6.67 � 10�11

(m/s2)(m)2

��N m2/kg2

(9.8 m/s2)(6.4 � 106 m)2

��(6.67 � 10�11 N m2/kg2)



Now let’s work on the units (carrying along the numerical value):

MEarth � 6.02 � 1024 �(m

(s)(2

m)(N

)2

m(k

2

g))2

�

� 6.02 � 1024 �ms2N

3 kmg

2

2

�.

Cancel the m2:

� 6.02 � 1024 �ms2

kNg2

�.

By definition 1 N � 1 kg m/s2. Substituting for N we have

� 6.02 � 1024 �ms2

kkgg

2

ms2

�.

Canceling as indicated, we are left simply with kg. So our final result is

MEarth � 6.02 � 1024 kg.

This is a lot of mass, and the Earth is only one small blue planet just 4000miles in radius! The value we have obtained agrees with the mass of theEarth obtained by other means, once again confirming Newton’s theory.

Newton’s Work: Impact and Reaction

Newton’s work opened whole new lines of investigation, both theoreticaland observational. In fact, much of our present science and also our tech-nology had their effective beginnings with the work of Newton and thosewho followed in his spirit. New models, new mathematical tools, and a newconfidence encouraged those followers to attack new problems, to opennew vistas of research, and to answer long-standing questions. The mod-ern view of science is that it is a continuing exploration of ever more in-teresting fields.

Newton’s influence was not limited to science alone. The period fol-lowing his death in 1727 was a period of further understanding and appli-cation of his discoveries and method. His influence was felt especially in

PART ONE 33


philosophy and literature, but also in many other fields outside science. Letus round out our view of Newton by referring to some of these effects.

The eighteenth century is often called the Age of Reason, the apogee ofthe so-called Enlightenment. “Reason” was the motto of the eighteenth-century philosophers. Enlightened by reason, especially scientific reason,humanity would overcome the darkness of ignorance and usher in a newage of the flowering of human potential. Such ideals appeared, for instance,in the following excerpt from Hymn to Science by the poet Mark Akenside(1721–1770).*

Science! thou fair effusive rayFrom the great source of mental day,

Free, generous, and refined!Descend with all thy treasures fraught,Illuminate each bewilder’d thought,

And bless my labouring mind. . . .

Oh! let thy powerful charms impartThe patient head, the candid heart,

Devoted to thy sway;Which no weak passions e’er mislead,Which still with dauntless steps proceed

Where reason points the way. . . .

Give me to learn each secret cause;Let Number’s, Figure’s, Motion’s laws

Reveal’d before me stand;These to great Nature’s scenes apply,And round the globe, and through the sky,

Disclose her working hand.

Many thinkers of the Enlightenment believed they could extend the tri-umph of human reason in science to other areas of human endeavor. As aresult, Newtonian physics, religious toleration, and republican governmentwere all advanced by the same movement. However, their theories aboutimproving religion and society were not convincingly connected. This doesnot mean there was really a logical link among these concepts. Nor weremany eighteenth-century thinkers, in any field or nation, much botheredby other gaps in logic and feeling. For example, they believed that “all men


* From Poems of Science, John Heath-Stubbs and Phillips Salman, eds. (New York: Penguin, 1984), pp. 150–152. We thank E.B. Sparberg for bringing this to our attention.


are created equal.” Yet they did little to remove the chains of black slaves,the ghetto walls imprisoning Jews, or the laws that denied rights to women.

Still, compared with the previous century, the dominant theme of theeighteenth century was moderation, the “happy medium.” The emphasis wason greater toleration of different opinions, restraint of excess, and balanceof opposing forces. Even reason was not allowed to question religious faithtoo strongly. Atheism, which some philosophers thought would logicallyresult from unlimited rationality, was still regarded with horror by mostEuropeans.

The Constitution of the United States of America is one of the most en-during achievements of this period. Its system of “checks and balances” wasdesigned specifically to prevent any one group from getting too muchpower. It attempted to establish in politics a state of equilibrium of oppos-ing trends. This equilibrium, some thought, resembled the balance betweenthe Sun’s gravitational pull and the tendency of a planet to fly off in astraight line. If the gravitational attraction upon the planet increased with-out a corresponding increase in planetary speed, the planet would fall intothe Sun. If the planet’s speed increased without a corresponding increasein gravitational attraction, it would escape from the solar system. Whenthe opposing tendencies balanced, harmony resulted.

Political philosophers, some of whom used Newtonian physics as amodel, hoped to create a similar balance in government. They tried to de-vise a system that would avoid the extremes of dictatorship and anarchy.According to James Wilson (1742–1798), who played a major role in writ-ing the American Constitution:

In government, the perfection of the whole depends on the balanceof the parts, and the balance of the parts consists in the indepen-dent exercise of their separate powers, and, when their powers areseparately exercised, then in their mutual influence and operationon one another. Each part acts and is acted upon, supports and issupported, regulates and is regulated by the rest.

Both Newton’s life and his writings seemed to support the idea of po-litical democracy. A former farm boy had attained the outermost reachesof the human imagination. What he had found there meant, first of all, thatthe same set of laws governed motion in the celestial and terrestrial spheres.This smashed the old beliefs about “natural place” and extended a newdemocracy throughout the Universe. Newton had shown that all matter,whether the Sun or an ordinary stone, was created equal; that is to say, allmatter had the same standing before “the Laws of Nature and of Nature’sGod.” (This phrase was used at the beginning of the Declaration of Inde-

PART ONE 35


pendence to justify the desire of the people in the American colonies tothrow off their oppressive political system and to become an independentpeople.) All political thought at this time was heavily influenced by New-tonian ideas. The Principia seemed to offer a parallel to theories aboutdemocracy. It seems logical that all people, like all natural objects, are cre-ated equal before nature’s creator.

In literature, too, as already indicated, many welcomed the new scien-tific viewpoint. It supplied new ideas, convenient figures of speech,metaphors, parallels, and concepts which writers used in poems and essays.Many poems of the eighteenth century referred to Newton’s discovery thatwhite light is composed of colors (see Chapter 8). Samuel Johnson advo-cated that words drawn from the vocabulary of the natural sciences be usedin literary works. He defined many such words in his Dictionary and illus-trated their application in his Rambler essays.

However, not everyone welcomed the new rational, scientific viewpoint.That viewpoint was based on the idea that nature consists only of mattermoving through empty space according to gravity and Newton’s laws ofmotion. Many writers and artists of the Romantic movement were partic-ularly disturbed by this so-called “mechanical world view” which, they ar-gued, replaced the vibrancy and beauty of nature with an ugly, lifeless worldof inert particles moving forever in empty space. Where in this system isthere room for the beauty and warmth and feeling of a gorgeous rainbow,a melodious concerto, or the emotions of love and hate, ambition and pride,happiness and sorrow?

Romanticism started in Germany about 1780 among young writers in-spired by the poet–philosopher Johann Wolfgang von Goethe. The mostfamiliar examples of Romanticism in English literature are the poems andnovels of Blake, Coleridge, Shelley, Byron, Scott, and Wordsworth. Mostof the Romantics scorned the mathematical view of nature. They believedthat any whole thing, whether a single human being or the entire Universe,is filled with a unique, nonmaterial spirit. This spirit cannot be explainedby reason; it can only be felt. The Romantics insisted that phenomena can-not be meaningfully analyzed and reduced to their separate parts by me-chanical explanations or pure reason alone. Contrast the following excerptfrom William Wordsworth’s (1770–1850) “The Tables Turned”* with Akenside’s “Hymn to Science” quoted earlier:*

Up! Up! my friend, and clear your looks,Why all this toil and trouble?


* ibid., p. 166.


Up! Up! my friend, and quit your books,Or surely you’ll grow double. . . .

Books! ’tis a dull and endless strife,Come, hear the woodland linnet,How sweet his music; on my lifeThere’s more of wisdom in it.

And hark! How blithe the throstle sings!And he is no mean preacher;Come forth into the light of things,Let Nature be your teacher.

The Romantic philosophers in Germany regarded Goethe as their great-est scientist as well as their greatest poet. They pointed in particular to histheory of color, which flatly contradicted Newton’s theory of light. Goetheheld that white light does not consist of a mixture of colors and that it isuseless to “reduce” or “torture” a beam of white light by passing it througha prism to study its separate spectral colors. Rather, he charged, the colorsof the spectrum are artificially produced in Newton’s experiment using theprism, acting on and changing the light which is itself pure.

In the judgment of all modern scientists on this point, Newton was rightand Goethe wrong. This does not mean that so-called Nature Philosophy, in-troduced by Friedrich Schelling in the early 1800s as the Romantic answerto Newtonian physics, was without any value. It encouraged speculation aboutideas, even if they were so general that they could not be easily tested by ex-periment. At the time, it was condemned by most scientists for just this rea-son. Today, most historians of science agree that Nature Philosophy eventu-ally played an important role in making possible certain scientific discoverieslater on. Among these was the general principle of conservation of energy,which is described in the next two chapters. This principle asserted that allthe “forces of nature,” that is, the phenomena of heat, gravity, electricity,magnetism, and so forth, are forms of one underlying “force” (which we nowcall energy). This idea had agreed well with the viewpoint of Nature Phi-losophy. But it also could be put eventually in a scientifically acceptable form.

Movements hostile to conventional science have in fact occurred fromtime to time since Antiquity, in various forms, and are again visible today.Some modern artists, some intellectuals, and most members of the “alter-native” or “new age” movements express deep-felt dislikes and mistrust ofscience. Their feelings are similar to, and historically related with, those ofthe Romantics. They are based in part on the mistaken notion that mod-ern scientists dogmatically claim to be able to find (or have) a mechanical

PART ONE 37


explanation for everything, whereas science is so powerful by being neitherdogmatic, nor beholden only to “mechanics,” nor ambitious to other fieldsin which it does not belong.

Even the Roman philosopher Lucretius (100–55 B.C.), who supportedthe atomic theory in his poem On the Nature of Things, wished to preservesome role for “free will” in the Universe, by suggesting that atoms mightswerve randomly in their paths. This was not enough for Romantics, oreven for some scientists. For example, Erasmus Darwin, a scientist andgrandfather of evolutionist Charles Darwin, asked:

Dull atheist, could a giddy danceOf atoms lawless hurl’d

Construct so wonderful, so wise,So harmonised a world?

The Romantic Nature philosophers thought they could discredit theNewtonian scientists by forcing them to answer this question. To say “yes,”they argued, would be absurd, and to say “no” would be disloyal to New-tonian beliefs. But the Newtonians succeeded quite well without commit-ting themselves to any definite answer to Erasmus Darwin’s question. Theywent on to discover immensely powerful and valuable laws of nature, whichare discussed in the chapters ahead.

Questions

1. Describe some of the impacts of Newton’s work outside the field of science.

2. What impact did Newtonian physics have on political thought?3. Why did some people eventually reject the new physics?4. Contrast the excerpts from the poems by Akenside and Wordsworth.5. The poem by Erasmus Darwin asks a question. What is it in your own

words? How did Nature Philosophers attempt to discredit Newtonianscientists?

CHAPTER 5. CONSERVING MATTER AND MOTION

An Example of Conservation of Momentum

(1) A space capsule at rest in space, far from the Sun or planets, has a massof 1000 kg. A meteorite with a mass of 0.1 kg moves toward it with a speed



of 1000 m/s. How fast does the capsule (with the meteorite stuck in it)move after being hit?

mA (mass of the meteorite) � 0.1 kg,

mB (mass of the capsule) � 1000 kg,

vA (initial speed of meteorite) � 1000 m/s,

vB (initial speed of capsule) � 0,

vA� (final speed of meteorite) � ?,

v�B (final speed of capsule) � ?.

The law of conservation of momentum states

mAvA � mBvB � mAv�A � mBv�B.

Inserting the values given, we have

(0.1 kg)(1000 m/s) � (1000 kg) (0)

� (0.1 kg) v�A � (1000 kg) v�B,

100 kg � m/s � (0.1 kg) v�A � (1000 kg) v�B.

Since the meteorite sticks to the capsule, v�B � v�A; so we can write

100 kg � m/s � (0.1 kg) v�A � (1000 kg) v�A,

100 kg � m/s � (1000 � 1 kg) v�A.

Therefore,

vA� � �10

1000

k0g.1

�

kmg/s

�

� 0.1 m/s

PART ONE 39


(in the original direction of the motion of the meteorite). Thus, the cap-sule (with the stuck meteorite) moves on with a speed of 0.1 m/s.

Another approach to the solution is to handle the symbols first, and sub-stitute the values as a final step. Substituting v�A for v�B and letting v�B � 0would leave the equation mAvA � mAv�A � mBv�B � (mA � mB)v. Solving forv�A we obtain

v�A � .

This equation holds true for any projectile hitting (and staying with) a bodyinitially at rest that moves on in a straight line after collision.

(2) An identical capsule at rest nearby is hit by a meteorite of the samemass as the other. However, this meteorite, hitting another part of the cap-sule, does not penetrate. Instead, it bounces straight back with almost nochange of speed. How fast does the capsule move after being hit? Since allthese motions are assumed to be along a straight line, we can drop the vec-tor notation from the symbols and indicate the reversal in direction of themeteorite with a minus sign.

The same symbols are appropriate as in (1):

mA � 0.1 kg, vB � 0,

mB � 1000 kg, v�A � 1000 m/s,

vA � 1000 m/s, v�B � ?.

The law of conservation of momentum states

mAvA � mBvB � mAv�A � mBv�B.

(1)

THUNK!

(2)

CLANG!

mAvA��(mA � mB)



Here,

(0.1 kg)(1000 m/s) � (1000 kg)(0),

� (0.1 kg)(�1000 m/s) � (1000 kg) v�B

100 kg � m/s � �100 kg � m/s � (1000 kg) v�B,

v�B � �20

100k0g0

�

kmg

/s� � 0.2 m/s.

Thus, the struck capsule moves on with about twice the speed of the cap-sule in (1). (A general symbolic approach to this solution can be taken, too.The result is valid only for the special case of a projectile rebounding per-fectly elastically from a body of much greater mass.)

There is a general lesson here. It follows from the law of conservation ofmomentum that a struck object is given less momentum if it absorbs the pro-jectile than if it reflects it. (A goalie who catches the soccer ball is pushedback less than one who lets the ball bounce off.) Some thought will help youto understand this idea: An interaction that merely stops the projectile is notas great as an interaction that first stops it and then propels it back again.

Doing Work on a Sled

Suppose a loaded sled of mass m is initially at rest on low-friction ice. You,wearing spiked shoes, exert a constant horizontal force F on the sled. Theweight of the sled is balanced by the upward push exerted by the ice, so Fis effectively the net force on the sled. You keep pushing, running faster andfaster as the sled accelerates, until the sled has moved a total distance d.

If the net force F is constant, the acceleration of the sled is constant.Two equations that apply to motion starting from rest with constant ac-celeration are

v � at

and

d � 1⁄2at2,

where a is the acceleration of the body, t is the time interval during whichit accelerates (i.e., the time interval during which a net force acts on thebody), v is the final speed of the body, and d is the distance it moves in thetime interval t.

PART ONE 41


(a) (b)

According to the first equation, t � v/a. If we substitute this expressionfor t in the second equation, we obtain

d � 1⁄2at2 � 1⁄2a�va2

2

� � 1⁄2 �va

2

�.

The work done on the sled is W � Fd. From Newton’s second law, F � ma, so

W � Fd

� ma � 1⁄2�va2

2

�.

The acceleration cancels out, giving

W � 1⁄2mv2.

Therefore, the work done in this case can be found from just the massof the body and its final speed. With more advanced mathematics, it canbe shown that the result is the same whether the force is constant or not.

More generally, we can show that the change in kinetic energy of a bodyalready moving is equal to the work done on the body. By the definitionof average speed

d � vavt.

If we consider a uniformly accelerated body whose speed changes from v0to v, the average speed (vav) during t is 1⁄2(v � v0). Thus,

d � �v �

2v0

� � t.

F

m


F

mV


By the definition of acceleration, a � � v/t; therefore, t � � v/a � (v � v0)/a.Substituting (v � v0)/a for t gives

d � �v �

2v0

� � �v �

av0

�

��(v � v0

2)(av � v0)�

� �v2 �

2av0

2

�.

The work W done is W � Fd, or, since F � ma:

W � ma � d

� ma �

� �m2

� (v2 � v20)

� 1⁄2 mv2 � 1⁄2 mv20.

CHAPTER 6. THE DYNAMICS OF HEAT

You will often see energies expressed in terms of other units. A few of themare listed here.

Unit name Symbol Definition Conversion

kilowatt hour kWhr A watt (W) is 1 J/s, so 1 kWh � 3.60 MJ1 J � 1 W � s. A kWh is the amount of energy delivered in 1 hr if 1 kJ is delivered persecond.

Calorie Cal The energy required to heat 4.19 kJ(or kilocalorie) (or kcal) 1 kg of water by 1°C.

British thermal Btu The energy required to heat 1.06 kJunit 0.454 kg by 0.556°C.

v2 � v20

�2a

PART ONE 43


Carnot’s Proof

Carnot’s proof of maximum efficiency of ideal, reversible engines starts withthe premise that when a cold object is in contact with a warmer one, the coldobject does not spontaneously cool itself further and so give more heat to thewarm object. However, an engine placed between the two bodies can moveheat from a cold object to a hot one. Thus, a refrigerator can cool a cold bot-tle further, ejecting heat into the hot room. You will see that this is not simple.Carnot proposed that during any such experiment, the net result cannot beonly the transfer of a given quantity of heat from a cold body to a hot one.

The engines considered in this case all work in cycles. At the end of eachcycle, the engine itself is back to where it started. During each cycle, it hastaken up and given off heat, and it has exerted forces and done work.

Consider an engine, labeled R in the figure, which suffers no internalfriction, loses no heat because of poor insulation, and runs so perfectly thatit can work backward in exactly the same way as forward (Figure A).

Now suppose someone claims to have invented an engine, labeled Z inthe next figure, which is even more efficient than the ideal engine R. Thatis, in one cycle it makes available the same amount of work, W, as the Rengine does, but takes less heat energy, Q�, from the hot object to do it(Q1� Q1). Since heat and energy are equivalent and since Q2 � Q1 � Wand Q2� � Q�1 � W, it will also be true that Q2� Q2 (Figure B).

Q2� � Q�1 � W,

Suppose the two engines are connected so that the work from one canbe used to drive the other. For example, the Z engine can be used to makethe R engine work like a refrigerator (Figure C).

B

ZQ′2

Q′1W

Hot

Cold

A

R

(a) (b)

Q2

Q1W

Hot object

Cold object

RQ2

Q1

Hot object

Cold object



At the end of one cycle, both Z and R are back where they started. Nowork has been done; the Z engine has transferred some heat to the coldobject; and the R engine has transferred some heat to the hot object. Thenet heat transferred is Q1 � Q�1, and the net heat taken from the cold ob-ject is Q2 � Q2. These are, in fact, the same

Q2 � Q2� � (Q1 � W ) � (Q�1 � W )

� Q1 � Q�1.

Because Z is supposed to be more efficient than R, this quantity shouldbe positive; that is, heat has been transferred from the cold object to thehot object. Nothing else has happened. But, according to the fundamentalpremise, this is impossible, and does not happen.

The only conclusion is that the Z engine was improperly “advertised”and that it is either impossible to build or that in actual operation it willturn out to be less efficient than R.

As for two different reversible engines, they must have the same effi-ciency. Suppose the efficiencies were different; then one would have to bemore efficient than the other. What happens when the more efficient en-gine is used to drive the other reversible engine as a refrigerator? The sameargument just used shows that heat would be transferred from a cold bodyto a hot one. This is impossible. Therefore, the two reversible engines musthave the same efficiency.

To actually compute that efficiency, you must know the properties of onereversible engine; all reversible engines working between the same tem-peratures must have that same efficiency. (Carnot computed the efficiencyof an engine that used an ideal gas instead of steam.)

CHAPTER 7. HEAT—A MATTER OF MOTION

Averages and Fluctuations

Molecules are too small, too numerous, and too fast for us to measure thespeed of any one molecule, its kinetic energy, or how far it moves before

RZ

C

Q2

Q1

Q ′2

Q ′1 W

Hot

Cold

PART ONE 45


colliding with another molecule. For this reason, the kinetic theory of gasesconcerns itself with making predictions about average values. The theoryenables us to predict quite precisely the average speed of the molecules ina sample of gas, the average kinetic energy, or the average distance the mol-ecules move between collisions.

Any measurement made on a sample of gas reflects the combined effectof billions of molecules, averaged over some interval of time. Such averagevalues measured at different times, or in different parts of the sample, willbe slightly different. We assume that the molecules are moving randomly.Thus, we can use the mathematical rules of statistics to estimate just howdifferent the averages are likely to be. We will call on two basic rules ofstatistics for random samples:

1. Large variations away from the average are less likely to occur than aresmall variations. (For example, if you toss 10 coins, you are less likelyto get 9 heads and 1 tail than to get 6 heads and 4 tails.)

2. Percentage variations are likely to be smaller for large samples. (Forexample, you are likely to get nearer to 50% heads by flipping 1000coins than by flipping just 10 coins.)

A simple statistical prediction is the statement that if a coin is tossedmany times, it will land “heads” 50% of the time and “tails” 50% of thetime. For small sets of tosses there will be many “fluctuations” (variations)to either side of the predicted average of 50% heads. Both statistical rulesare evident in the charts. The top chart shows the percentage of heads insets of 30 tosses each. Each of the 10 black squares represents a set of 30tosses. Its position along the horizontal scale indicates the percentage of

0 10

30-Toss sets

20 30 40 50 60 70 80 90 100 % Heads

0 10

90-Toss sets

20 30 40 50 60 70 80 90 100 % Heads

0 10

180-Toss sets

20 30 40 50 60 70 80 90 100 % Heads

(a)

(b)

(c)



heads. As we would expect from Rule 1, there are more values near the the-oretical 50% than far from it. The second chart is similar to the first, buthere each square represents a set of 90 tosses. As before, there are morevalues near 50% than far from it. And, as we would expect from Rule 2,there are fewer values far from 50% than in the first chart.

The third chart is similar to the first two, but now each square repre-sents a set of 180 tosses. Large fluctuations from 50% are less common stillthan for the smaller sets.

Statistical theory shows that the average fluctuation from 50% shrinks inproportion to the square root of the number of tosses. We can use this ruleto compare the average fluctuation for sets of, say, 30,000,000 tosses withthe average fluctuation for sets of 30 tosses. The 30,000,000-toss sets have1,000,000 times as many tosses as the 30-toss sets. Thus, their average fluc-tuation in percent of “heads” should be 1,000 times smaller!

These same principles hold for fluctuations from average values of anyrandomly distributed quantities, such as molecular speed or distance be-tween collisions. Since even a small bubble of air contains about a quintil-lion (1018) molecules, fluctuations in the average value for any isolated sam-ple of gas are not likely to be large enough to be measurable. A measurablylarge fluctuation is not impossible, but extremely unlikely.

Deriving an Expression for Pressure from the Kinetic Theory

We begin with the model of a gas described in Section 7.2: “a large num-ber of very small particles in rapid, disordered motion.” We can assumehere that the particles are points with vanishingly small size, so that colli-sions between them can be ignored. If the particles did have finite size, theresults of the calculation would be slightly different. But the approxima-tion used here is accurate enough for most purposes.

The motions of particles moving in all directions with many different ve-locities are too complex as a starting point for a model. So we fix our at-tention first on one particle that is simply bouncing back and forth betweentwo opposite walls of a box. Hardly any molecules in a real gas would ac-tually move like this. But we will begin here in this simple way and later inthis chapter extend the argument to include other motions. This later partof the argument will require that one of the walls be movable. Therefore,we will arrange for that wall to be movable, but to fit snugly into the box.

In Chapter 5, you saw how the laws of conservation of momentum andenergy apply to cases like this. When a very light particle hits a more mas-sive object, like the wall, very little kinetic energy is transferred. If the col-lision is elastic, the particle will reverse its direction with very little change

PART ONE 47


in speed. In fact, if a force on the outside of the wall keeps it stationaryagainst the impact from inside, the wall will not move during the collisions.Thus no work is done on it, and the particles rebound without any changein speed.

How large a force will these particles exert on the wall when they hit it?By Newton’s third law the average force acting on the wall is equal and op-posite to the average force with which the wall acts on the particles. Theforce on each particle is equal to the product of its mass times its acceler-ation (F � ma), by Newton’s second law. The force can also be written as

F � ,

where �mv is the change in momentum. Thus, to find the average forceacting on the wall we need to find the change in momentum per seconddue to molecule–wall collisions.

Imagine that a particle, moving with speed vx (the component of v inthe x direction) is about to collide with the wall at the right. The compo-nent of the particle’s momentum in the x direction is mvx. Since the parti-cle collides elastically with the wall, it rebounds with the same speed. There-fore, the momentum in the x direction after the collision is m(�vx). Thechange in the momentum of the particle as a result of this collision is

final momentum � initial momentum � change in momentum,

(�mvx) � (mvx) � (�2mvx).

Vx

−Vx

�(mv)�

�t

L

L

L

m

v



Note that all the vector quantities considered in this derivation have onlytwo possible directions: to the right or to the left. We can therefore indi-cate direction by using a � or a � sign, respectively.

Now think of a single particle of mass m moving in a cubical containerof volume L3 as shown in the figure.

The time between collisions of one particle with the right-hand wall is thetime required to cover a distance 2L at a speed of vx; that is, 2L/vx. If 2L/vxequals the time between collisions, then vx/2L equals the number of colli-sions per second. Thus, the change in momentum per second is given by

� � � � � � � �,

(�2mvx) � �2vLx� � .

The net force equals the rate of change of momentum. Thus, the aver-age force acting on the molecule (due to the wall) is equal to �mvx

2/L, andby Newton’s third law, the average force acting on the wall (due to the mol-ecule) is equal to �mvx

2/L. So the average pressure on the wall due to thecollisions made by one molecule moving with speed vx is

P � �AF

� � �LF

2� � �mLv3

2x

� � ,

where V (here L3) is the volume of the cubical container.Actually, there are not one but N molecules in the container. They do

not all have the same speed, but we need only the average speed in order

mv2x

�V

�mv2x

�L

change in momentumper second

number of collisionsper second

change in momentumin one collision

vmVx

L

L

m=

2

AverageForce

−mvx mvx

−2 mvx

PART ONE 49


to find the pressure they exert. More precisely, we need the average of thesquare of their speeds in the x direction. We call this quantity (vx

2)av. Thepressure on the wall due to N molecules will be N times the pressure dueto one molecule, or

P � .

In a real gas, the molecules will be moving in all directions, not just inthe x direction; that is, a molecule moving with speed v will have threecomponents: vx, vy, and vz. If the motion is random, then there is no preferred direction of motion for a large collection of molecules, and(v2

x)av � (v2y )av � (v2

z)av. It can be shown from Pythagoras’ theorem that v2 � v2

x � v2y � v2

z. These last two expressions can be combined to give

(v2)av � 3(v2x )av

or

(v2x ) � 1/3(v2)av.

By substituting this expression for (v2x )av in the pressure formula, we get

P ��Nm �

V1/3(v2)av�

� 1/3 �NVm� (v2)av.

Notice now that Nm is the total mass of the gas, and therefore Nm/V isjust the density D. So

P � 1/3D(v2)av.

This is our theoretical expression for the pressure P exerted on a wall bya gas in terms of its density D and the molecular speed v.

Vx

V2

Vy

V

Nm(v2x)av

�V



CHAPTER 8. WAVE MOTION

Calculating the Wavelength from anInterference Pattern

V � (S1S2) � separation between S1 and S2. (S1 and S2 may be actual sourcesthat are in phase, or two slits through which a previously prepared wavefront passes.)

l � OQ � distance from sources to a far-off line or screen placedparallel to the two sources,

x � distance from center axis to point P along the detectionline,

L � OP � distance to point P on detection line measured fromsources.

Waves reaching P from S1 have traveled farther than waves reaching P fromS2. If the extra distance is � (or 2�, 3�, etc.), the waves will arrive at P inphase. Then P will be a point of strong wave disturbance. If the extra dis-tance is 1⁄2� (or 3⁄2�, 5⁄2�, etc.), the waves will arrive out of phase. Then Pwill be a point of weak or no wave disturbance.

With P as center, draw an arc of a circle of radius PS2; it is indicated onthe figure by the dotted line S2M. Then line segment PS2 equals line seg-ment PM. Therefore, the extra distance that the wave from S travels toreach P is the length of the segment SM.

PART ONE 51

s2

L

L

s1

do

dM

M

PX

X

Q


Now if d is very small compared to l, as you can easily arrange in prac-tice, the circular arc S2M will be a very small piece of a large-diameter circle, or nearly a straight line. Also, the angle S1MS2 is very nearly 90°.Thus, the triangle S1S2/M can be regarded as a right triangle. Furthermore,angle S1S2/M is equal to angle POQ. Then the right triangle S1S2M is sim-ilar to triangle POQ:

�SS

1

1

MS2� � �

OXP� or �

S1

dM� � �

XL

�.

If the distance l is large compared to x, the distances l and L are nearlyequal. Therefore,

�S1

dM� �

But S1/M is the extra distance traveled by the wave from source S1. For Pto be a point of maximum wave disturbance, S1/M must be equal to n�(where n � 0 if P is at Q, and n � 1 if P is at the first maximum of wavedisturbance found to one side of Q, etc.). So the equation becomes

�nd��

xl�

and

� � �dnxl�.

This important result says that if you measure the source separation d, thedistance l, and the distance x from the central line to a wave disturbancemaximum, you can calculate the wavelength �.

The Sonic Boom

In the last half century a new kind of noise has appeared: the sonic boom.An explosion-like sonic boom is produced whenever an object travelsthrough air at a speed greater than the speed of sound (supersonic speed).Sound travels in air at about 340 m/s. Many types of military airplanes cantravel at two or three times this speed. Flying at such speeds, the planes

x�l



unavoidably and continually produce sonic booms, which can cause phys-ical damage, and anxiety in people and animals. SST (Supersonic Trans-port) planes such as the Concorde are now in civilian use in some countries.The unavoidable boom raises important questions. What are the conse-quences of this technological “progress”? Who gains, and what fraction ofthe population do they represent? Who and how many pay the price? Mustwe pay it; must SST’s be used? How much say has the citizen in decisionsthat affect the environment so violently?

The formation of a sonic boom is similar to the formation of a wake by aboat. Consider a simple point source of waves. If the source remains in thesame position in a medium, the wave it produces spreads out symmetricallyaround it, as in Diagram 1. If the source of the disturbance is moving throughthe medium, each new crest starts from a different point, as in Diagram 2.

Notice that the wavelength has become shorter in front of the objectand longer behind it. This is called the Doppler effect. The Doppler effectis the reason that the sound an object makes seems to have a higher pitchwhen it is moving toward you and a lower pitch when it is moving awayfrom you. In Diagram 3, the source is moving through the medium fasterthan the wave speed. Thus, the crests and the corresponding troughs over-lap and interfere with one another. The interference is mostly destructiveeverywhere except on the line tangent to the wave fronts, indicated in Di-agram 4. The result is a wake that spreads like a wedge away from the mov-ing source, as in the diagram.

All these concepts apply not only to water waves but also to sound waves,including those disturbances set up in air by a moving plane as it pushesthe air out of the way. If the source of sound is moving faster than the speedof sound wave, then there is a cone-shaped wake (in three dimensions) thatspreads away from the source.

Actually, two cones of sharp pressure change are formed. One cone orig-inates at the front of the airplane and one at the rear, as indicated in thegraph at the right.

4

31 2

PART ONE 53


Because the double shock wave follows along behind the airplane, theregion on the ground where people and houses may be struck by the boom(the “sonic-boom carpet”) is as long as the supersonic flight path itself. Insuch an area, typically thousands of kilometers long and 80 km wide, theremay be millions of people. Tests made with airplanes flying at supersonicspeed have shown that a single such cross-country flight by a 315-ton su-personic transport plane would break many thousands of dollars worth ofwindows, plaster walls, etc., and cause fright and annoyance to millions ofpeople. Thus, the supersonic flight of such planes has been confined toover-ocean use. It may even turn out that the annoyance to people on ship-board, on islands, and on coastal areas near the flight paths is so great thatover-ocean flights, too, will have to be restricted.


normal

AirPressure

milliseconds0 10 20 30 40 50 60 70 80 90 100 110

A

B

C


Model, Analogy, Hypothesis, Theory

Model, analogy, hypothesis, and theory have similar but distinct meaningswhen applied to physics. An analogy is a corresponding situation which,though perhaps totally unrelated to the situation at hand, helps you un-derstand it. Many electronic circuits have analogs in mechanical systems.A model is a corresponding situation that may offer a picture of what “is re-ally going on” and therefore can be taken more seriously as an explanation.An electron rotating around a nucleus is one model for an atom. A hypoth-esis is a statement that can usually be directly or indirectly tested. ToFranklin, the statement “lightning is caused by electricity” was at first a hy-pothesis. A theory is a more general construction, perhaps putting togetherseveral models and hypotheses to explain a collection of effects that previ-ously seemed unrelated. Newton’s explanation of Kepler’s laws, Galileo’sexperiments in mechanics and, finally, the Cavendish experiment were all part of the theory of universal gravitation. This is a good example of atheory.

A well-tested theory, such as Newton’s theory of gravitation or Einstein’stheory of relativity, is a robust part of science, explaining a myriad of indi-vidual events or facts, and not to be confused with the vernacular use of“just a theory.”

CHAPTER 9. EINSTEIN AND RELATIVITY THEORY

Differences in Speed for Light Waves TravelingParallel and Perpendicular to the Ether Wind

Instead of light waves moving parallel and perpendicular to the ether wind,we examine an equivalent situation: a swimmer swimming at constant speed,first parallel and perpendicular to a current, in a river 1 mi wide. Assumethe swimmer can swim at 2 mi/hr and the stream runs 1 mi/hr from leftto right in the diagram below. We will calculate the time required for theswimmer to travel 1 mi each way with and against the current and 1 miback and forth across the current.

With and against the current

1 mile

v = 1 mi/hr

PART ONE 55


Traveling 1 mi with the current, the swimmer’s speed is enhanced by thespeed of the current, while traveling 1 mi back against the current, his speedis hindered by it. Thus the total time for the round trip is

t � �

� �3

1m

mi/h

ir

� �

� 1.33 hr

Across the current

In order to swim directly across the river from the starting point and back,the swimmer, in each direction, must head toward a point upstream from thedestination point. The path taken relative to the fixed shore will be directlyacross and back, 1 mi in each direction. But the path taken in each directionby the swimmer relative to the flowing water will be along the hypothenuse ofa right triangle formed by the 1-mi width of the river and the speed of theriver current times one-half the total time for the round trip (1 mi/hr)t/2.

Using the Pythagorean theorem, the total distance traveled by the swim-mer at the speed of 2 mi/hr is

(2 mi/hr) t � 2�(1 mi)�2 � [(1� mi/hr�) t/2]2�.�

In order to solve for t, cancel 2 on both sides and square both sides to get

mi2/hr2 t2 � mi2 � 1⁄4 mi2/hr2 t2.

Cancel mi2, multiply through by hr2 and solve for t2:

3⁄4t2 � 1 hr2,

t � �4/3� hr � 1.15 hr.

1 mile

(1 mi/hr) t/2

(2 mi/hr) t/2

1 mile v = 1 mi/hr

1 mi�1 mi/hr

1 mi��2 mi/hr � 1 mi/hr

1 mi��2 mi/hr � 1 mi/hr



Note that the time to cross the river back and forth at constant swimmingis less than the time it took in the earlier example to swim the same dis-tance parallel to the current and back.

Michelson and Morley reasoned that exactly the same kind of resultwould occur for a light beam split in half—one-half sent “swimming” per-pendicular to the supposed ether wind and back, the other half “swim-ming” parallel to the wind and back. Although the two halves of the beamstarted out together, the one sent parallel to the wind should return slightlybehind the one sent parallel to the wind. The difference in time was ex-pected to be small but detectable. Yet, when comparing the two light wavesexperimentally, they could find no difference in the time of travel of the twobeams. We now know from Einstein’s second postulate that the times had to be the same, and that the ether model, while usually appealing, ismisleading.

PART ONE 57


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